Given the two quadratic equations:

(I) $$2x^2 - 13x + 20 = 0$$

(II) $$y^2 + 10y + 24 = 0$$

Solve both equations and determine the correct relationship between $x$ and $y$ from the options below.

Question

Given the two quadratic equations:

(I) $$2x^2 - 13x + 20 = 0$$

(II) $$y^2 + 10y + 24 = 0$$

Solve both equations and determine the correct relationship between $x$ and $y$ from the options below.

Accepted Answer

**Solution:**

1. Solve the first quadratic equation:

$$ 2x^2 - 13x + 20 = 0 $$

Using the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$, where $a=2$, $b=-13$, $c=20$:

$$
$\Delta = (-13)^2 - 4 \times 2 \times 20 = 169 - 160 = 9$
$$

$$
$x = \frac{13 \pm 3}{4}$
$$

So,

- $$x_1 = \frac{13 + 3}{4} = \frac{16}{4} = 4$$
- $$x_2 = \frac{13 - 3}{4} = \frac{10}{4} = 2.5$$

2. Solve the second quadratic equation:

$$ $y^2 + 10y + 24 = 0$ $$

Here, $a=1$, $b=10$, $c=24$:

$$
$\Delta = 10^2 - 4 \times 1 \times 24 = 100 - 96 = 4$
$$

$$
$y = \frac{-10 \pm 2}{2}$
$$

So,

- $$y_1 = \frac{-10 + 2}{2} = \frac{-8}{2} = -4$$
- $$y_2 = \frac{-10 - 2}{2} = \frac{-12}{2} = -6$$

3. Compare the roots:

- $x$ values are $4$ and $2.5$
- $y$ values are $-4$ and $-6$

Since both $x$ roots are greater than both $y$ roots, the relationship is:

$$ x > y $$

**Answer:** If $x > y$

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